1. 13 part 2 Enzyme kinetics
酵素動力學
溫鳳君0993b303
姜喆云0993b039

2. 心智圖

3. 課程重點 part 2
Kinetics are the branch of science concerned with the 反應速率.
Enzyme kinetics are to determine the 最大反應速率and binding affinities for substrates and inhibitors.

4. 1.一級反應(First-order reaction) :unimolecular reaction A→P V=-[ A]/dt= k[A] 2.二級反應(second-order reaction)= Bimolecular reaction A + B →P + Q或2 A → P + Q 註:很少三級反應

5. Transition state是在energy profile最高點
Increase chemical reaction rate
(a)升高溫度 fromT1 to T2
(b) 加催化劑
Transition state是在energy profile最高點
The reaction rate is 按比例 to the
concentration of reactant molecules with the
transition-state energy

6. Free energy of activation(G‡)
the energy required to raise the average energy of 1 mol of reactant to the transition state energy
Decreasing G‡ increases the reaction rate
Enzyme did change kinetics!!!!!!!!

7. The Michaelis-Menten Equation
假設形成ES complex
假設ES complex is in rapid equilibrium with free enzyme
Breakdown of ES to form products is
assumed to be slower than
1) formation of ES and
2) Breakdown of ES to re-form E and S
E+S ES E+P

8. Km 根據M-M assumption，假設E+S ES E+P
根據Briggs & Haldane assumed that Vf = Vd
under steady-state!
推導出
Km= Sum of dissociation rate constants
ES synthesis rate constant

9. Km’s meaning
Km和ES binding capacity 有關
Km和反應速率有關
當最大速率的一半時，Km會和基質濃度相同

10. kcat A measure of catalytic activity
the activity of one molecule of enzyme
當M-M model fits, k2 = kcat = Vmax/Et

11. The Ratio kcat/Km Define the catalytic efficiency of enzyme
The upper limit for kcat/Km is the diffusion limit -the rate at which E and S diffuse together

12. 問題 1.
Under what conditions can a bisubstrate enzyme-catalyzed reaction with a double-displacement mechanism be treated kinetically as if it is a single-substrate reaction?    (I.e., under what conditions does its Michaelis-Menten equation approximate that for a single-substrate reaction?) A:
The Michaelis Menten equation for a bisubstrate reaqction with a double displacement mechanism reduces to the equation for a single substrate reaction when the concentration of the second substrate is extremely large relative to the binding affinity for that substrate, as is the case when that substrate is water.

13. 2. Explain mathmematically how a value for Km can be obtained from the Vo vs
So graph when Vo = 1/2 Vmax.
When Vo = Vmax/2, then Vmax/2 = Vmax
So ,
Km + So
cancelling Vmax,
1/2 = So or Km + So = 2So
Km + So
or Km = So at Vo = (value of) Vmax/2